资源描述
课时规范练29等比数列及其前n项和基础巩固组1.已知等比数列an满足a1=14,a3a5=4(a4-1),则a2=() A.2B.1C.12D.182.在正项等比数列an中,a2,a48是方程2x2-7x+6=0的两个根,则a1a2a25a48a49的值为()A.212B.93C.93D.353.(2017安徽黄山市二模)已知数列an的前n项和为Sn,且a1=2,an+1=Sn+1(nN*),则S5=()A.31B.42C.37D.474.设首项为1,公比为23的等比数列an的前n项和为Sn,则()A.Sn=2an-1B.Sn=3an-2C.Sn=4-3anD.Sn=3-2an5.(2017全国)等差数列an的首项为1,公差不为0.若a2,a3,a6成等比数列,则an前6项的和为()A.-24B.-3C.3D.86.设等比数列an的前n项和为Sn.若S2=3,S4=15,则S6=()A.31B.32C.63D.647.设数列an是首项为a1,公差为-1的等差数列,Sn为其前n项和.若S1,S2,S4成等比数列,则a1的值为.8.(2017北京)若等差数列an和等比数列bn满足a1=b1=-1,a4=b4=8,则a2b2=.9.(2017江苏,9)等比数列an的各项均为实数,其前n项和为Sn.已知S3=74,S6=634,则a8=.10.(2017河南新乡二模,文17)在数列an中,a1=12,an的前n项和Sn满足Sn+1-Sn=12n+1(nN*).(1)求数列an的通项公式an以及前n项和Sn;(2)若S1+S2,S1+S3,m(S2+S3)成等差数列,求实数m的值.导学号24190754综合提升组11.(2017四川广元二诊)已知数列an的前n项和为Sn,且对任意正整数n都有an=34Sn+2成立.若bn=log2an,则b1 008=()A.2 017B.2 016C.2 015D.2 01412.设等比数列an满足a1+a3=10,a2+a4=5,则a1a2a3an的最大值为.13.已知an是公差为3的等差数列,数列bn满足b1=1,b2=13,anbn+1+bn+1=nbn.(1)求an的通项公式;(2)求bn的前n项和.创新应用组14.已知数列an的前n项和为Sn,在数列bn中,b1=a1,bn=an-an-1(n2),且an+Sn=n.(1)设cn=an-1,求证:cn是等比数列;(2)求数列bn的通项公式.答案：1.Ca3a5=4(a4-1),a42=4(a4-1),解得a4=2.又a4=a1q3,且a1=14,q=2,a2=a1q=12.2.Ba2,a48是方程2x2-7x+6=0的两个根,a2a48=3.又a1a49=a2a48=a252=3,a250,a1a2a25a48a49=a255=93.3.Dan+1=Sn+1(nN*),Sn+1-Sn=Sn+1(nN*),Sn+1+1=2(Sn+1)(nN*),数列Sn+1是首项为3,公比为2的等比数列.则S5+1=324,解得S5=47.4.DSn=a1(1-qn)1-q=a1-anq1-q=1-23an1-23=3-2an,故选D.5.A设等差数列的公差为d,则d0,a32=a2a6,即(1+2d)2=(1+d)(1+5d),解得d=-2,所以S6=61+652(-2)=-24,故选A.6.CS2=3,S4=15,由等比数列前n项和的性质,得S2,S4-S2,S6-S4成等比数列,(S4-S2)2=S2(S6-S4),即(15-3)2=3(S6-15),解得S6=63,故选C.7.-12由已知得S1=a1,S2=a1+a2=2a1-1,S4=4a1+432(-1)=4a1-6,而S1,S2,S4成等比数列,(2a1-1)2=a1(4a1-6),整理,得2a1+1=0,解得a1=-12.8.1设等差数列an的公差为d,等比数列bn的公比为q,由题意知-1+3d=-q3=8,即-1+3d=8,-q3=8,解得d=3,q=-2.故a2b2=-1+3-1(-2)=1.9.32设该等比数列的公比为q,则S6-S3=634-74=14,即a4+a5+a6=14.S3=74,a1+a2+a3=74.由得(a1+a2+a3)q3=14,q3=1474=8,即q=2.a1+2a1+4a1=74,a1=14,a8=a1q7=1427=32.10.解 (1)an+1=Sn+1-Sn=12n+1,当n2时,an=12n.又a1=12,当n=1时上式也成立.an=12n,Sn=121-12n1-12=1-12n.(2)由(1)可得:S1=12,S2=34,S3=78.S1+S2,S1+S3,m(S2+S3)成等差数列,12+34+m34+78=212+78,解得m=1213.11.A在an=34Sn+2中,令n=1得a1=8,an=34Sn+2成立,an+1=34Sn+1+2成立,两式相减得an+1-an=34an+1,an+1=4an,又a10,数列an为等比数列,an=84n-1=22n+1,bn=log2an=2n+1,b1 008=2 017,故选A.12.64由已知a1+a3=10,a2+a4=(a1+a3)q=5,得q=12,所以a1=8,所以a1a2a3an=8n121+2+(n-1)=2-12n2+7n2,所以当n=3或n=4时,a1a2a3an取最大值为2-1232+732=26=64.13.解 (1)由已知,得a1b2+b2=b1,因为b1=1,b2=13,所以a1=2.所以数列an是首项为2,公差为3的等差数列,通项公式为an=3n-1.(2)由(1)和anbn+1+bn+1=nbn,得bn+1=bn3,因此bn是首项为1,公比为13的等比数列.记bn的前n项和为Sn,则Sn=1-13n1-13=32-123n-1.14.(1)证明 an+Sn=n,an+1+Sn+1=n+1.-得an+1-an+an+1=1,2an+1=an+1,2(an+1-1)=an-1,an+1-1an-1=12,an-1是等比数列.又a1+a1=1,a1=12,首项c1=a1-1,c1=-12,公比q=12.又cn=an-1,cn是以-12为首项,以12为公比的等比数列.(2)解 由(1)可知cn=-1212n-1=-12n,an=cn+1=1-12n.当n2时,bn=an-an-1=1-12n-1-12n-1=12n-1-12n=12n.又b1=a1=12代入上式也符合,bn=12n.
展开阅读全文